PROPERTIES OF THE DIFFUSION EQUATION

3.2 PROPERTIES OF THE DIFFUSION EQUATION

In cases where the tracer mass is neither created nor destroyed, such as occurs in chemi- cal and biochemical reactions, the contaminant is called conservative, and S m ⫽ 0 in Equation 3.16. In this case, the transport of the contaminant is described by

∂ c 3 ∂ c 3 ∂ 2 c ᎏ ᎏ⫹ V i ᎏ ᎏ⫽ D ᎏ ᎏ

∂ t i⫽ 冱 1 ∂ x i i⫽ 冱 1 i ∂ x 2 i

Consider now a change of variables from x i and t to x⬘ i and t, where the new variables x⬘ i are de fined by

(3.18) where V i is a constant mean velocity in the x i -direction, and the x⬘ i -coordinate measures

x⬘ i ⫽x i ⫺V i t

locations relative to the mean position of the tracer particles, given by V i t . The derivatives in the (x i , t) space are related to the derivatives in the (x⬘ i , t) space by the following rela- tions derived from the chain rule,

t j⫽ 冱 1 ∂ x j ⬘ ∂ t ∂ t

where (·) represents any scalar function of x and t. Combining Equations 3.18 to 3.20 yields

冱 V j ᎏ ᎏ⫹ᎏ ᎏ

j⫽ 1 ∂ x j ⬘

Substituting Equations 3.21 and 3.22 into the advection–di ffusion equation, Equation 3.17, yields the transformed equation in (x⬘ i , t)-space,

ᎏ ∂ ᎏ⫽ c 冱 D i ᎏ ∂ ᎏ c

t i⫽ 1 ∂ i ′ 2

96 FATE AND TRANSPORT IN AQUATIC SYSTEMS

which is more commonly written in the Cartesian form

2 ᎏ ∂ c ᎏ⫽D ᎏ ∂ ᎏ⫹D c ᎏ ∂ 2 c ∂ ᎏ⫹D 2 x c y

Equation 3.24 occurs widely in engineering applications and is generally referred to as the di ffusion equation. This equation has been studied in detail throughout the mathematical literature, particularly in the context of heat conduction, and analytical solutions for a wide variety of initial and boundary conditions are available in many textbooks. Using these solutions together with the transformation given by Equation 3.18 provides many useful results to describe the mixing process when the mean flow is steady and spatially uniform.

3.2.1 Fundamental Solution in One Dimension

Consider the case where the tracer is distributed uniformly in the y and z directions and di ffusion occurs only in the x-direction. The diffusion equation is then given by

∂ c ∂ 2 c ᎏ ᎏ⫽D ᎏ ᎏ

If a tracer of mass M is introduced instantaneously at x ⫽ 0 at time t ⫽ 0 (well mixed over y and z), and tracer concentrations are always equal to zero at x ⫽ ⫾ ∞, the initial and boundary conditions are given by

c (x, 0) ⫽ ᎏ ᎏ δ(x)

and

(3.27) where A is the area in the yz plane over which the contaminant is well mixed, and δ(x) is

c (⫾ ∞, t) ⫽ 0

the Dirac delta function, which is de fined by

x⫽ 0 冕

δ (x) dx ⫽ 1 (3.28)

A graph of the Dirac delta function, centered at x 0 (where x 0 ⫽ 0 in Equation 3.28), is illus- trated in Figure 3.3. The solution to Equation 3.25 subject to initial and boundary condi- tions given by Equations 3.26 and 3.27 can be obtained using the Fourier transform,

f (k, t), de fined as

f (k, t) ⫽ ᎏ ᎏ 冕 c (x, t)e ikx dx

and the inverse Fourier transform de fined as

f (k, t)e 冕 ⫺ikx dk

c (x, t) ⫽ ᎏ ᎏ

PROPERTIES OF THE DIFFUSION EQUATION

(x)

FIGURE 3.3 Dirac delta function.

In applying the Fourier transform to solve Equation 3.25, it is useful to note that integra- tion by parts yields the identity

dx ⫽ ⫺ ᎏ ᎏ 冕 c (x, t)e dx ⫽ ⫺k f (k, t) (3.31)

Multiplying Equation 3.25 by

e i kx

and integrating yields the homogeneous ordinary di fferential equation

ᎏ ᎏ⫹k 2 D

x f⫽ 0 (3.32)

and the initial condition derived from Equation 3.26 is

The solution to Equation 3.32 subject to the initial condition given by Equation 3.33 can easily be shown to be

This is the solution of the di ffusion equation, Equation 3.25, in the Fourier-transformed (k, t)-space, and therefore the solution in the (x, t)-space, c(x, t), can be obtained by

98 FATE AND TRANSPORT IN AQUATIC SYSTEMS

x FIGURE 3.4 One-dimensional di ffusion.

substituting Equation 3.34 into the inverse Fourier transform relation, Equation 3.30, to yield

冢 4D x t 冣

c (x, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ

A 兹 4 苶π苶D 苶 苶 x t

This result indicates that the concentration distribution resulting from the instantaneous introduction of a mass M is in the form of a Gaussian distribution with a variance growing with time, as illustrated by a plot of Equation 3.35 given in Figure 3.4. To verify that the concentration distribution given by Equation 3.35 is Gaussian, consider the equation for a Gaussian distribution given by

where µ is the mean of the distribution, σ is the standard deviation, and M 0 is the total area under the curve. Note that a normal distribution is the same as a Gaussian distribution, except that the area under the curve, M 0 , is equal to unity. Comparing the fundamental solution of the di ffusion equation, Equation 3.35, to the Gaussian distribution, Equation 3.36, it is clear that the fundamental solution is Gaussian with the mean and standard deviation given by

(3.38) This result demonstrates that a mass of contaminant released instantaneously into a stagnant

σ⫽ 兹2苶D 苶 x 苶 t

fluid will attain a concentration distribution that is Gaussian, with a maximum concentration remaining at the location the mass was released, and the standard deviation of the distribu-

tion grows in proportion to the square root of the elapsed time since the release.

PROPERTIES OF THE DIFFUSION EQUATION

The one-dimensional advection–di ffusion equation in a moving fluid is given by

x ᎏ ∂ ᎏ ∂x 2 x (3.39) where V is the fluid velocity in the x-direction. Equation 3.39 transforms to

ᎏ ᎏ⫹Vᎏ ᎏ⫽D

∂ c ∂ 2 c ᎏ ᎏ⫽D x ᎏ ᎏ

in the x⬘–t domain, where x⬘ ⫽ x ⫺ Vt. The initial and boundary conditions corresponding to an instantaneous release at x ⫽ 0 and t ⫽ 0 with boundaries in finitely far away from the release location are

c (x⬘, 0) ⫽ ᎏ ᎏ δ (x⬘)

and

(3.42) where A is the area in the yz plane over which the contaminant is well mixed. The solution

c (⫾ ∞, t) ⫽ 0

to Equation 3.40 is the same as the fundamental solution for a stationary fluid and is there- fore given by

c (x⬘, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ

A 兹 4 苶π苶D 苶 x 苶 t

which in the x–t domain is given by

(x ⫺ V t ) 2

c (x, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ

A 兹 4 苶π苶D 苶 x 苶 t

The concentration distribution described by Equation 3.44 and illustrated in Figure 3.5 describes the mixing of a tracer released instantaneously into a flowing fluid, where the tracer undergoes one-dimensional di ffusion. If the fluid is stagnant, V ⫽ 0 and the resulting con- centration distribution is symmetrical around x ⫽ 0 and is described by Equation 3.35.

Example 3.1 One hundred kilograms of a contaminant is spilled into a small river and instantaneously mixes across the entire cross section of the river. The cross section of the river is approximately trapezoidal in shape, with a bottom width of 5 m, side slopes of 2:1

t) , c (x

0 Vt

FIGURE 3.5 Solution to a one-dimensional advection–di ffusion equation. (From Chin, David A., Water-Resources Engineering. Copyright © 2000. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, NJ.)

FATE AND TRANSPORT IN AQUATIC SYSTEMS

(H:V), and a depth of flow of 3 m. The discharge in the river is estimated as 30 m 3 /s, and the dispersion coe

fficient for mixing along the river is estimated as 10 m 2 /s. Estimate (a) when the maximum contaminant concentration will be observed at a park recreation area

10 km downstream of the spill, and (b) the maximum concentration expected at the park. (c) If a safe level of this contaminant in recreational waters is 10 µg/L, how long after the spill can the park expect to resume normal operations?

SOLUTION (a) From the data given, M ⫽ 100 kg, D x ⫽ 10 m 2 /s, and the flow rate, Q, in

the river is 30 m 3 /s. The cross-sectional area, A, of the river is given by

A ⫽ by ⫹ my 2

where b ⫽ 5 m, y ⫽ 3 m, and m ⫽ 2; hence,

A⫽ 5(3) ⫹ 2(3) 2 ⫽ 33 m 2

and

Q 3 0 V⫽ᎏ ᎏ⫽ᎏ ᎏ ⫽ 0.909 m/s

The distance, x m , of the maximum concentration from the spill location at any time, t, is given by

x m ⫽ Vt

Therefore, for x m ⫽ 10 km ⫽ 10,000 m,

x m 1 0 , 0 0 0 t⫽ᎏ ᎏ⫽ᎏ ᎏ ⫽11,000 s ⫽ 3.06 h

Hence, the park can expect to see the peak contaminant concentration 3.06 h after the spill occurs.

(b) The maximum contaminant concentration at any time (t) and location (x) is given by Equation 3.44 for x ⫽ Vt as

A 兹 4 苶π苶D 苶 苶 x t

c (x, t) ⫽ ᎏ ᎏ⫽ ᎏᎏᎏ ⫽ 2.58 ⫻ 10 ⫺3 kg/m 3 ⫽ 2.58 mg/L

Hence, the maximum contaminant concentration observed at the recreation area is expected to be 2.58 mg/L.

(c) When the concentration at the recreation area is 10 µg/L ⫽ 10 ⫺5 kg/m 3 , Equation

3.44 requires that

(x ⫺ V t ) 2

c (x, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ

A 兹 4 苶π苶D 苶 x 苶 t

冤 4 ( 10 )t 冥

1 0 0 (10,00 0 ⫺ 0 .909t) 2

10 ⫽ᎏ ᎏ exp ⫺ᎏ ᎏ

33 兹 苶 4 π 苶 苶0苶)t (1 苶

PROPERTIES OF THE DIFFUSION EQUATION

which yields t ⫽ 9400 s and 12,850 s. Clearly, the concentration is above 10 µg/L from t⫽ 9400 to 12,850 s, and the park water is expected to be safe when t ⬎ 12,850 s ⫽ 3.57 h after the spill.

In general, water-quality measurements would be required to ensure the safety of the water at the recreational area prior to resuming normal operations.

3.2.2 Principle of Superposition

The principle of superposition states that if a homogeneous di fferential equation is linear and there are several solutions that satisfy the equation, the sum of the solutions also satis fies the equation, and this sum is called the total solution of the linear differential equation. Additionally, if the boundary and initial conditions of the individual solutions are also linear, the boundary and initial conditions of the total solution are equal to the sum of the boundary and initial conditions of the individual solutions. To illustrate the principle

of superposition, consider that c 1 and c 2 are separate solutions to the one-dimensional di ffusion equation; therefore,

∂ c x ᎏ ᎏ 2 x 2 ∂ (3.46) t ∂

ᎏ ᎏ⫽D 2

Adding these equations yields

ᎏ ᎏ⫹ᎏ ᎏ⫽D 2 x 1 ᎏ ᎏ ⫹D ᎏ ᎏ 2 (3.47)

and invoking the linearity property leads to

which demonstrates that the sum of the solutions to the di ffusion equation (i.e., c 1 ⫹c 2 ) is also a solution to the di ffusion equation. To demonstrate the effect of linearity on the boundary conditions of the total solution, suppose that the boundary conditions on c 1 and

Then clearly the boundary condition of c 1 ⫹c 2 is (c 1 ⫹c 2 )( ∞, t) ⫽ 0

FATE AND TRANSPORT IN AQUATIC SYSTEMS

Similarly, if the initial conditions corresponding to instantaneous releases of masses M 1

A the initial condition of c 1 ⫹c 2 is

(c 1 ⫹c 2 )(x, 0) ⫽ ᎏ ᎏ 1 δ(x ⫺ x

A 1 )⫹ᎏ ᎏ δ(x ⫺ x 2 )

This result demonstrates that the principle of superposition can be used to determine the contaminant distribution resulting from two simultaneous mass inputs at two di fferent locations, based on the contaminant distribution resulting from mass releases at single locations. Several additional applications of the principle of superposition are given in the following sections.

Distributed Source Consider the initial spatial distribution of a contaminant given in Figure 3.6, where the initial one-dimensional concentration distribution is given by

(3.55) This initial concentration distribution is equivalent to an in finite number of adjacent

c (x, 0) ⫽ f (x)

instantaneous sources of mass f (x) A dx located along the x-axis between x L and x R , where

A is the cross-sectional area over which the contaminant is well mixed. For each of these

f (x)

Initial condition: c (x,0) = f (x)

f( ξ)

FIGURE 3.6 Initial concentration distribution.

PROPERTIES OF THE DIFFUSION EQUATION

f (x )

Initial condition

FIGURE 3.7 Step-function initial condition.

incremental sources located a distance ξ away from the origin, the fundamental solution of the one-dimensional di ffusion equation is applicable, and the resulting concentration distribution is given by

Using the principle of superposition to sum the solutions for all the incremental sources, results in the total solution

c (x, t) ⫽ 冕 ᎏ ᎏ exp ⫺ᎏ ᎏ

A frequently encountered initial concentration distribution is the step function shown in Figure 3.7, which is described by

c (x, 0) ⫽ f (x) ⫽ 0 xⱕ 0 冦 (3.58)

0 x⬎ 0

and substituting this initial condition into Equation 3.57 yields

c (x, t) ⫽ 冕 ᎏ 0 ᎏ exp ⫺ᎏ ᎏ

(x ⫺ ξ ) 2

Changing variables from x to u, where

x⫺ ξ u⫽ᎏ ᎏ

4 苶D 苶 x 苶 t

FATE AND TRANSPORT IN AQUATIC SYSTEMS

Equation 3.59 becomes

π 冕 x / 兹4苶D 苶 xt 苶

c (x, t) ⫽ ᎏ 0 ᎏ 2 e ⫺u du

This integral cannot be evaluated analytically, but is similar to a special function in math- ematics called the error function, erf(z), which is de fined as

erf(z) ⫽ ᎏ ᎏ

⫺ ξ d ξ 兹 (3.62) 苶

and values of this function are tabulated in Appendix E.1. It is useful to note the property

(3.63) and the limits

erf(⫺z) ⫽ ⫺erf(z)

(3.64) Comparing the solution of the di ffusion equation, Equation 3.61, with the definition of the

erf(0) ⫽ 0 and erf( ∞) ⫽ 1

error function, Equation 3.62, leads to

c ∞ 2 x / 兹4苶D 苶 xt 苶 2

c (x, t) ⫽ ᎏ ᎏ 0 e ⫺u du ⫺ 冕 e ⫺u du 兹 π 苶 冤冕 0 0 冥

π 苶 冤 2 2 冢 兹4苶 苶 D 苶 x t 冣冥

2 冤 冢 兹4苶 D 苶 x 苶 t 冣冥

A further simpli fication of Equation 3.65 comes from the definition of the complementary error function , erfc(z), where

(3.66) The solution given by Equation 3.65 can therefore be written in the form

erfc(z) ⫽ 1 ⫺ erf(z)

2 冢 兹4苶 苶 D 苶 x t 冣

c (x, t) ⫽ ᎏ ᎏ erfc 0 ᎏ ᎏ

and the concentration distribution, c(x, t), is illustrated in Figure 3.8. The corresponding solution for a fluid moving with a velocity V is given by

2 冢 兹 苶D 4 苶 x 苶 t 冣

c (x, t) ⫽ ᎏ ᎏ erfc 0 ᎏ ᎏ

c x⫺ V t

This concentration distribution is identical to that illustrated in Figure 3.8, with the excep- tion that the origin moves with the fluid at a velocity V.

PROPERTIES OF THE DIFFUSION EQUATION

c(x,t )= c 0 x 2 erfc − 4D x t

Initial condition

FIGURE 3.8 Di ffusion from a step-function initial condition.

Example 3.2

A long drainage canal is gated at the downstream end and is designed to retain the runo ff from an agricultural area. The runoff into the canal is expected to infiltrate into the ground water. After a severe storm, the concentration of a toxic pesticide in the channel rises to 5 mg/L and is distributed uniformly throughout the channel. Because of the threat of flooding, the gate at the downstream end of the canal is opened and water in the canal flows downstream at a velocity of 20 cm/s. (a) If the longitudinal dispersion

coe fficient in the canal is 5 m 2 /s, give an expression for the concentration as a function of time at a location 400 m downstream of the gate. (b) How long after the gate is opened will the concentration at the downstream location be equal to 1 mg/L?

SOLUTION (a) From the data given, c 0 ⫽ 5 mg/L ⫽ 0.005 kg/m 3 , V ⫽ 20 cm/s ⫽ 0.20 m/s, and D x ⫽5m 2 /s. At x ⫽ 400 m, Equation 3.68 gives the concentration as a function of time as

2 冢 兹 4 苶D 苶 苶 x t 冣

c (x, t) ⫽ ᎏ ᎏ erfc 0 ᎏ ᎏ

c x⫺ V t

冢 兹 苶( 4 苶 5 ) 苶 t 冣

0.0 05 40 0⫺ 0 . 2 0t

8.94 ⫺ 0 .0447t

c (400, t) ⫽ ᎏ ᎏ erfc ᎏ ᎏ ⫽ 0.0025 erfc ᎏ ᎏ 2

(b) When the concentration 400 m downstream of the gate is equal to 1 mg/L ⫽ 0.001 kg/m 3 ,

8.94 ⫺ 0 .0447t

0.001 ⫽ 0.0025 erfc ᎏ ᎏ

which leads to

8.94 ⫺ 0 .0447t

erfc ᎏ ᎏ ⫽ 0.4

FATE AND TRANSPORT IN AQUATIC SYSTEMS

Using the error function tabulated in Appendix E.1,

8.94 ⫺ 0 .0447t ᎏ ᎏ ⫽ 0.595 兹

which leads to

t⫽

80.5 s

Therefore, the concentration 400 m downstream of the gate will reach 1 mg/L approxi- mately 81 s after the gate is opened.

Transient Source Suppose that a contaminant source is located along the x-axis, and . the source injects the contaminant with a time-varying mass flux, m (t), uniformly over a . cross-sectional area A in a stagnant fluid. This scenario is equivalent to a mass of m (t) dt

being released during every consecutive time interval dt. The concentration distribution,

dc (x, t), resulting from an instantaneous mass of m ( τ) dτ released from x ⫽ 0 at time τ is given by

冤 4D x ( t ⫺ τ) 冥

and superimposing all of the resulting concentration distributions yields

c (x, t) ⫽ ᎏ ᎏ exp d 冕 ⫺ᎏ

冤 4D x ( t ⫺ τ) 冥

x ᎏ (3.70)

0 A 兹 苶 4 苶 π 苶 D 苶 x ( t⫺ 苶τ) 苶

This equation describes the concentration distribution resulting from a transient source at x ⫽ 0. In the case of a spatially distributed transient source with mass flux m·(x, t), the principle of superposition indicates that the resulting concentration distribution is given by

c (x, t) ⫽ 冕 冕 x ᎏ ᎏ exp ⫺ᎏ ᎏ ␰)

m ( ␰ , τ )d ␰ d τ

(x ⫺ 2

冤 4D x (t ⫺ τ) 冥

0 L A 兹 苶 4 π 苶 D 苶 x 苶 ( t⫺ 苶 τ) 苶

where x L and x R are the upper and lower bounds of the tracer source location.

Impermeable Boundaries In the superposition examples cited so far, the boundary conditions have required that the tracer concentration approaches zero as x approaches in finity. Therefore, the superimposed concentration distributions all have boundary condi- tions in which the tracer concentration approaches zero as x approaches in finity. In cases where impermeable boundaries exist in relatively close proximity to the tracer source, the di ffusion equation must satisfy boundary conditions that require zero mass flux across the

PROPERTIES OF THE DIFFUSION EQUATION

c(x,t )

c 2 (x,t )

Impermeable boundary

Real source with boundary, c 1 +c 2

Real source Image source without

without boundary, c 1

boundary, c 2

FIGURE 3.9 Impermeable boundary condition.

. impermeable boundary. This scenario is illustrated in Figure 3.9. Since mass flux, M x , is

governed by the Fickian-type di ffusion equation

M x ⫽ ⫺D x ᎏ ᎏ c (3.72)

an impermeable boundary requires that the concentration gradient, ∂c/∂x, be equal to zero at the boundary. Referring to Figure 3.9, if the tracer source is located at x ⫽ 0 and the imper- meable boundary is located at x ⫽ L, then if an identical source is superimposed at x ⫽ 2L, the resulting (superimposed) concentration distribution has the following properties within the domain x ∈(⫺∞, L]: (1) satisfies the diffusion equation (via linearity); (2) satisfies the ini- tial condition of an instantaneous mass release at x ⫽ 0; and (3) ∂c/∂x ⫽ 0 at x ⫽ L, due to the symmetry of the superimposed solutions around x ⫽ L. These results demonstrate that the symmetrical placement of an image source produces a solution that satis fies the diffusion equation as well as the required initial and boundary conditions. Applying this result to the case of an instantaneous source of mass, M, located at a distance L from an impermeable boundary, indicates that the resulting concentration distribution is given by

⫺ᎏ x 2 ⫺

A 兹 4 苶π苶D 苶 x 苶 t 冤 冢 4D x t 冣 冢 4 D x t 冣冥

ᎏ ⫹ exp ⫺ᎏ (x ᎏ 2 L ) (3.73)

c (x, t) ⫽ ᎏ

exp

Example 3.3 One kilogram of a contaminant is spilled into an open channel at a location

50 m from the end of the channel. The channel has a rectangular cross section 10 m wide and

2 m deep, and the dispersion coe fficient along the channel is estimated as 10 m 2 /s. (a) Assuming that the contaminant is initially well mixed across the channel, express the concentration as a function of time at the end of the channel. (b) How long will it take for the contaminant con- centrations 25 m upstream (in the direction of the channel end) to be 10% higher than the con- centration 25 m downstream of the spill (in the direction away from the channel end)?

SOLUTION (a) The concentration distribution is given by

A 兹 苶π苶D 4 苶 苶 x t 冤 冢 4D x t 冣 冢 4 D x t 冣冥

c (x, t) ⫽ ᎏ ᎏ exp ⫺ᎏ x ᎏ ⫹ exp ⫺ᎏ (x ᎏ 2 L )

FATE AND TRANSPORT IN AQUATIC SYSTEMS

where the x-coordinate is measured from the spill location in the direction of the end of the channel. From the data given, M ⫽ 1 kg, A ⫽ 10 m ⫻ 2 m ⫽ 20 m 2 ,D

2 x ⫽ 10 m /s, L ⫽ 50 m, and x ⫽ 50 m (at end of channel). The concentration as a function of time at the end of the

channel is therefore given by

20 兹4苶 苶(1 π 苶0苶)t 苶 冤 冢 4( 1 0 )t 冣 冢 4 (1 0 ) t 冣冥

c (50, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ ⫹ exp ⫺ᎏ ᎏ

which reduces to

c (50, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ 兹 苶

(b) The concentration 25 m upstream from the spill (x ⫽ 25 m) is given by

20 兹4苶 苶(1 π 苶0苶)t 苶 冤 冢 4( 1 0 )t 冣 冢 4 (1 0 ) t 冣冥

c (25, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ ⫹ exp ⫺ᎏ ᎏ

t 苶 冤 冢 t 冣 冢 t 冣冥

and the concentration 25 m downstream from the spill (x ⫽ ⫺25 m) is given by

20 兹4苶 苶(1 π 苶0苶)t 苶 冤 冢 4( 1 0 )t 冣 冢 4 (10 ) t 冣冥

c (⫺25, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ ⫹ exp ⫺ᎏ ᎏ

t 冤 冢 t 冣 冢 t 冣冥

When c(25, t) is 10% higher than c(⫺25, t), then

exp(⫺15.6/t) ⫹ exp (⫺141/t) ᎏᎏᎏᎏ ⫽ 1.1 exp (⫺15.6/t) ⫹ exp (⫺391/t)

Solving this equation for t yields t ⫽ 55 s. Therefore, after 55 s the concentrations 25 m upstream and downstream of the spill di ffer by 10%.

3.2.3 Solutions in Higher Dimensions

Solutions to the one-dimensional di ffusion equation shown so far have been intended to demonstrate some useful analytical procedures for obtaining solutions to relatively compli- cated problems. Many of these solutions are obtained by superimposing the fundamental solution of the one-dimensional di ffusion equation. These analytical procedures are also applicable in higher dimensions, as illustrated in the following sections.

PROPERTIES OF THE DIFFUSION EQUATION

Two-Dimensional Diffusion The fundamental di ffusion problem in two dimensions is given by

ᎏ ∂ c 2 ᎏ⫽ D 2 ᎏ ∂ ᎏ⫹D c x 2 ᎏ y ∂ ᎏ c (3.74)

with initial and boundary conditions

c (x, y, 0) ⫽ ᎏ ᎏ δ (x, y)

(3.76) where M is the mass of contaminant injected, L is the length over which the mass is uni-

c (⫾ ∞, ⫾∞, t) ⫽ 0

formly distributed in the z-direction, and δ (x, y) is the two-dimensional Dirac delta func- tion de fined by

⫹ ∞ ⫹ δ (x, y) ⫽ ∞

冦 0, otherwise

δ (x) dx dy ⫽ 1 (3.77)

The solution to the fundamental two-dimensional di ffusion problem is given by (Carslaw and Jaeger, 1959)

冢 4D x t 4D y t 冣

The principle of superposition can be applied to the fundamental two-dimensional solution of the di ffusion equation to yield the concentration distribution, c(x, y, t), resulting from an initial mass distribution, g(x, y), as

c (x, y, t) ⫽ 冕 x ᎏ ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ

x 2 y 2 g ( ξ , η )d ξ d η

(x ⫺ ξ ) 2 (y ⫺ η ) 2

1 冕 y 1 4 π tL 兹 苶 D x 苶 D 苶 y 冤 4 D x t 4 D y t 冥

where the contaminant source is located in the region x ∈ [x 1 ,x 2 ], y ∈[y 1 ,y 2 ]. An example of di ffusion from a two-dimensional rectangular source is shown in Figure 3.10, where taking

D x ⫽D y leads to a symmetrical di ffusion pattern. Superposition in time can also be applied to . yield the concentration distribution, c(x, y, t), resulting from a continuous mass input m (t) as

c (x, y, t) ⫽ 冕

ᎏ exp d τ

2 ⫺ᎏ x ᎏ⫺ᎏ y ᎏ

0 4 (3.80) π(t ⫺ τ ) L 兹 苶 D 苶 x D 苶 y 冤 4D x ( t ⫺ τ) 4D y ( t ⫺ τ) 冥

where the transient source is located at x ⫽ 0, y ⫽ 0. In the case of a distributed transient . source, m (x, y, t), the resulting concentration distribution, c(x, y, t), is given by

c (x, y, t) ⫽ 冕

x 2 y 2 m ·( ξ , η, t ) d ξd η d t

(x ⫺ ξ) 2 ( y ⫺ η) 2

冤 4D x (t ⫺ τ ) 4 D y (t ⫺ τ ) 冥

ᎏ ᎏ exp ⫺ᎏ

1 4 π ( t⫺ τ ) L 兹 苶 D x 苶 D y 苶

FATE AND TRANSPORT IN AQUATIC SYSTEMS

t=t 1 t=t 2 t=t 3

t=t 4 t=t 5 t=t 6

FIGURE 3.10 Two-dimensional di ffusion from a finite source.

Example 3.4 One kilogram of a contaminant is spilled at a point in a 4-m-deep reservoir and is instantaneously mixed over the entire depth. (a) If the di ffusion coefficients in the

N-S and E-W directions are 5 and 10 m 2 /s, respectively, calculate the concentration as a function of time at locations 100 m north and 100 m east of the spill. (b) What is the con- centration at the spill location after 5 min?

SOLUTION (a) The concentration distribution is given by

冢 4D x t 4D y t 冣

c (x, y, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ

4 πtL 兹 D 苶 x 苶 D y 苶

From the data given, M ⫽ 1 kg, L ⫽ 4 m, D x ⫽5m 2 /s (N-S), and D y ⫽ 10 m 2 /s (E-W). At 100 m north of the spill, x ⫽ 0 m, y ⫽ 100 m, and the concentration as a function of time is given by

c (0, 100, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ ⫽ᎏ ᎏ exp ⫺ᎏ ᎏ kg/m 3

4 πt(4) 兹 5 苶( 苶1苶0苶)苶

At 100 m east of the spill, x ⫽ 100 m, y ⫽ 0 m, and the concentration as a function of time is given by

冢 4 ( 1 0 )t 冣 t

c (100, 0, t) ⫽ ᎏ ᎏ exp ⫺ᎏ ᎏ ⫽ᎏ ᎏ exp ⫺ᎏ ᎏ kg/m 3

4 πt(4) 兹 苶( 5 苶1苶0苶)苶

PROPERTIES OF THE DIFFUSION EQUATION

(b) At the spill location, x ⫽ 0 m and y ⫽ 0 m, and the concentration as a function of time is given by

c (0, 0, 300) ⫽ ᎏ ᎏ ⫽ 9.37 ⫻ 10 ⫺6 kg/m 3 ⫽ 9.37 µg/L

Therefore, the concentration at the spill location after 5 min is 9.37 µg/L.

Example 3.5

A vertical di ffuser discharges industrial wastewater at a rate of 5 m 3 /s uni- formly over a 5-m-deep reservoir. If the wastewater contains 50 mg/L of a toxic contami-

nant for 24 h, calculate the concentration of the contaminant as a function of time during this 24-h period at a distance of 100 m from the outfall. Assume that the di ffusion coe

fficient in the reservoir is equal to 10 m 2 /s. SOLUTION The concentration as a function of time is given by

c (x, y, t) ⫽ 冕 ᎏ ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ

冤 4D x ( t ⫺ τ) 4D y ( t ⫺ τ) 冥

0 4 π (t⫺ τ ) L 兹 苶 D x 苶 D y 苶

In this case, the two-dimensional contaminant mass flux, m , is

. m ⫽ Qc ⫽ 5(50 ⫻ 10 ⫺3 ) ⫽ 0. 25 kg/s

Since L ⫽ 5 m, D x ⫽D y ⫽ 10 m 2 /s, and at 100 m from the outfall x 2 ⫹y 2 ⫽ 100 2 m 2 , the con- centration 100 m from the outfall, c 100 (t), is given by

冤 4(10 ) ( t⫺ τ) 冥

c 100 (t) ⫽ 冕 ᎏ ᎏ exp ⫺ᎏ ᎏ d τ

0 4 π(t ⫺ τ )( 5)(10)

⫽ 0.000398 冕 ᎏ ᎏ exp ⫺ᎏ ᎏ d τ

This equation can be integrated numerically to yield the magnitude of the concentration 100 m from the outfall as a function of time. It is interesting to note that Equation 3.82 can also be expressed in terms of the well function, W(u), commonly used in ground-water applications, and tabulated values or approximations to W(u) can be used to aid integration.

Three-Dimensional Diffusion The fundamental di ffusion problem in three dimen- sions is given by

ᎏ ᎏ⫽D x ᎏ ᎏ⫹D ᎏ ᎏ⫹D ᎏ

FATE AND TRANSPORT IN AQUATIC SYSTEMS

with initial and boundary conditions

c (x, y, z, 0) ⫽ M δ(x, y, z)

(3.85) where δ (x, y, z) is the three-dimensional Dirac delta function defined by

c (⫾ ∞, ⫾∞, ⫾∞, t) ⫽ 0

δ(x, y, z) ⫽ ∞,

冦 0, otherwise

x⫽ 0, y ⫽ 0, z ⫽ 0

and

δ(x) dx dy dz ⫽ 1 (3.86)

The solution to the fundamental three-dimensional di ffusion problem is given by (Carslaw and Jaeger, 1959)

冢 4D x t 4D y t 4D z t 冣

The principle of superposition can be applied to the fundamental three-dimensional solu- tion of the di ffusion equation to yield the concentration distribution, c(x, y, z, t), resulting from an initial mass distribution (per unit volume), g(x, y, z), as

c (x, y, z, t) ⫽ 冕 x 冕 y 冕 z ᎏ 3/ 2 ᎏ exp ⫺ᎏ ᎏ⫺ ᎏ ᎏ⫺ ᎏ ᎏ

(3.88) where the contaminant source is located in the region x ∈ [x 1 ,x 2 ], y ∈ [y 1 ,y 2 ], z ∈ [z 1 ,z 2 ].

Superposition in time can also be applied to yield the concentration distribution, c(x, y, z, t),

resulting from a continuous mass input m (t) as

. m ( τ) dτ ᎏᎏᎏ

c (x, y, z, t) ⫽ 冕 0 [4 π(t⫺τ)] 3/2

x 苶 D 苶 exp y D 苶 z 冤 4D x ( t ⫺ τ) 4D y ( t ⫺ τ) 4D z ( t ⫺ τ) 冥

⫺ᎏ ᎏ⫺ᎏ ᎏ⫺ᎏ 兹D ᎏ 苶

(3.89) where the transient source is located at x ⫽ 0, y ⫽ 0, z ⫽ 0. In the case of a distributed transient

. source, m (x, y, z, t), the resulting concentration distribution, c(x, y, z, t), is given by

ᎏ ζ d c τ (x, y, z, t) ⫽ 冕

ᎏ ( ξ , η, ᎏᎏ ζ , τ ) d ξ d η d

x 1 冕 y 1 冕 z 1 [ 4 π ( t⫺ τ )] 3 /2 兹 苶 D x 苶 D y 苶 D z 苶

⫻ exp ⫺ᎏ ( x ⫺ 2 (y ᎏ⫺ᎏ ⫺ ξ) ᎏ⫺ᎏ η) 2 ( z ⫺ ᎏ ζ ) 冤 2

4 D z (t ⫺ τ ) 冥

4 D x (t ⫺ τ) 4D y (t ⫺ τ )

Example 3.6 One kilogram of a toxic contaminant is released deep into the ocean and spreads in all three coordinate directions. The N-S, E-W, and vertical di ffusion coefficients

are 10, 15, and 0.1 m 2 /s, respectively. (a) Find the concentration at a point 100 m north,

PROPERTIES OF THE DIFFUSION EQUATION

100 m east, and 10 m above the release point as a function of time. (b) What is the con- centration at the release point after 24 h?

SOLUTION (a) The concentration as a function of time is given by

冢 4D x t 4D y t 4D z t 冣

c (x, y, z, t) ⫽ ᎏ 3/2 ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ⫺ᎏ ᎏ

(4 πt) 兹 D 苶 x 苶 D 苶 y D 苶 z

In this case, M ⫽ 1 kg, D ⫽ 10 m 2 /s, D ⫽ 15 m x 2 y /s, D z ⫽ 0.1 m 2 /s, and therefore the con- centration as a function of time at x ⫽ 100 m, y ⫽ 100 m, z ⫽ 10 m, is given by

冤 4 (1 0 ) t 4 (1 5 ) t 4( 0 .1 )t 冥

c (100, 100, 10, t ) ⫽ ᎏᎏᎏᎏ 3/2 ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ⫺ᎏ ᎏ

(4 πt) 兹1 苶0苶( 苶1苶5苶)( 苶0苶.1 苶)苶

t 3/ 2 冢 t 冣

⫽ᎏ ᎏ exp ⫺ᎏ ᎏ

kg/m 3

(b) The concentration, c 0 , at the release point, x ⫽ 0, y ⫽ 0, z ⫽ 0, is given by

and at t ⫽ 24 h ⫽ 86,400 s the concentration at the release point, c 0 , is given by

c ⫽ᎏ ᎏ ⫽ 2.29 ⫻ 10 ⫺10

kg/m ⫽ 2.29 ⫻ 10 ⫺4 µg/L

Concentrations at this level would not be detectable, and the contaminant has dissipated, for all practical purposes.

Example 3.7 Ten kilograms of a contaminant in the form of a 1 m ⫻ 2 m ⫻ 2 m par- allepiped is released into the deep ocean. The N-S, E-W, and vertical di ffusion coe

fficients are 10, 5, and 0.05 m 2 /s, respectively. Find the concentration as a function of time at a location 50 m north, 50 m east, and 10 m above the centroid of the initial mass

release. SOLUTION The concentration distribution is given by

c (x, y, z, t) ⫽ 冕 x ᎏ ᎏ exp ⫺ᎏ ᎏ⫺ᎏ ᎏ⫺ᎏ ᎏ

x 2 y 2 z 2 g ( ξ, η , ζ )d ξ d η d ζ

(x ⫺ ξ ) 2 (y ⫺ η ) 2 (z ⫺ ζ ) 2

1 冕 y 1 冕 z 1 ( 4 π t ) 3/2 兹 苶 D 苶 x D y 苶 D 苶 z 冤 4 D x t 4 D y t 4 D z t 冥

where the initial concentration distribution, g( ξ, η, ζ ), is given by

⫺ 0.5 ⱕ ξ ⱕ 0.5, ⫺1.0 ⱕ η ⱕ 1.0, ⫺1.0 ⱕ ζ ⱕ 1.0 g ( ξ,η,ζ) ⫽ (1)( 2 )( 2 )m

ᎏ 1 0 ᎏ⫽ 2.5 kg/m k g 3 3 ,

otherwise

FATE AND TRANSPORT IN AQUATIC SYSTEMS

In this case, D ⫽ 10 m 2 x /s, D y ⫽5m 2 /s, and D z ⫽ 0.05 m 2 /s, and at x ⫽ 50 m, y ⫽ 50 m, z⫽ 10 m, the concentration as a function of time is given by

c (50, 50, 10, t) ⫽ ᎏᎏ 3/2 ᎏ (4 πt) 兹1苶0苶(5 苶)( 苶0苶.0 苶5苶)苶苶

冤 4 (1 0 )t 4 ( 5) t 4 ( 0 .05 ) t 冥

冤 4 0t

0 .2t 冥

⫽ᎏ 3 /2 ᎏ 冕 冕 冕 exp ⫺ᎏ ᎏ⫺ᎏ ᎏ⫺ᎏ ᎏ d ξ dη dζ

2 0t

This integral can be evaluated numerically to determine the values of c(50, 50,10, t) for speci fied values of t.

3.2.4 Moment Property of the Diffusion Equation

Consider the Cartesian form of the di ffusion equation given by Equation 3.24. Multiplying this equation by x⬘ 2 and integrating over x⬘ between ⫾ ∞ yields

2 z 冕 x⬘ ᎏ ⫺ ᎏ dx⬘ (3.91) ∞ ∂ t ⫺ ∞ ∂ x ⬘ ⫺ ∞ ∂ y ′ ⫺ ∞ ∂ z ⬘ 2

2 冕 x⬘ ᎏ ᎏ dx⬘⫹D

ᎏ ᎏx⬘ 2 dx⬘⫽ D x

x⬘ 2

ᎏ ᎏ dx⬘⫹ D 2 y

To evaluate these integrals, assume that the tracer concentrations are equal to zero at x⬘⫽ ⫾ ∞, which means that

(3.93) Applying these conditions to Equation 3.91 and integrating by parts yields ∂ ∞

c⫽ 0 at x⬘ i ⫽⫾ ∞

ᎏ ᎏ 冕 x⬘ 2 c dx⬘ ⫽ 2D y ᎏ 2 冕 x⬘ x 2 冕 c dx⬘⫹ D ᎏ c dx⬘⫹ D z ᎏ ᎏ x⬘ 2 c dx⬘ (3.94)

Integrating Equation 3.94 with respect to y⬘ from ⫺ ∞ to ⫹∞, applying Equations 3.92 and

3.93, and simplifying yields ∂ ∞ ∞

x⬘ c dx⬘dy⬘ ⫽ 2D x 冕 冕 c dx⬘dy⬘⫹ D z ᎏ ᎏ x⬘ c dx⬘dy⬘ (3.95)

Integrating Equation 3.95 with respect to z⬘ from ⫺ ∞ to ⫹∞, applying Equations 3.92 and

3.93, and simplifying yields ∂ ∞ ∞ ∞

ᎏ ᎏ 冕 冕 冕 x⬘ 2 c dx⬘dy⬘dz⬘ ⫽ 2D

x 冕 冕 冕 c dx⬘dy⬘ dz⬘

PROPERTIES OF THE DIFFUSION EQUATION

The integral term on the right-hand side of Equation 3.96 is equal to the total mass, M, of the tracer, where

M⫽ 冕

c dx⬘dy⬘dz⬘

and for conservative tracers M is a constant for the duration of the di ffusion process. Equation 3.96 can therefore be written in the form

x⬘ 2 c dx⬘dy⬘dz⬘ ⫽ 2D

where the partial derivative with respect to time has been replaced by the total derivative with respect to time, since the quantity being di

fferentiated depends only on time. 2 The variance of the concentration distribution along the x⬘-axis, σ 2 x⬘ , is given by

σ 2 ⫽ᎏ ᎏ x⬘ x⬘ 2 c dx⬘dy⬘dz⬘

and the integrated di ffusion equation, Equation 3.98, can therefore be written in the form

This rather remarkable result indicates that in a uniform flow field the diffusion coefficient,

D x , is equal to one-half the rate of growth of variance, σ 2 x⬘ , regardless of the initial condi- tions. Similar results are obtained for D y and D z by multiplying the original di ffusion equa- tion by y⬘ 2 and z⬘ 2 prior to integration, yielding

The practical utility of these equations, which relate the di ffusion coefficients to the variances of the tracer-concentration distributions, is that the variances can be meas- ured using tracer studies, and then the di ffusion coefficients are equal to one-half of the rate of growth of the respective variances. These measured (and validated) di ffusion coe fficients can then be used in the analysis and design of systems to control contami- nant transport.

2 The spatial dimensions have been removed by integration.

FATE AND TRANSPORT IN AQUATIC SYSTEMS

TABLE 3.1 Data for Example 3.8

Time, t (h)

TABLE 3.2 Results for Example 3.8

Time, t (h)

A 10-kg slug of Rhodamine WT dye is released into the ocean, and the con- centration distribution of the dye is measured every 3 h for the 12-h duration of daylight when the dye can be seen. The horizontal variances of the dye cloud as a function of time is given in Table 3.1. Estimate the horizontal di ffusion coefficients.

SOLUTION In accordance with Equations 3.101 and 3.102, the di ffusion coefficients can be approximated by

Therefore, between t ⫽ 0 and t ⫽ 3 h,

These di ffusion coefficients can be taken as the approximate values at t ⫽ (0 ⫹ 3)/2 ⫽ 1.5 h. Repeating this analysis for subsequent time intervals, the di ffusion coefficients as a func- tion of time are given in Table 3.2. Note that the di ffusion coefficients are clearly not steady and are increasing with time. As the plume expands, it experiences a wider variation of ocean currents; hence, mixing occurs at a more rapid rate with increasing time.

3.2.5 Nondimensional Form

Consider the general advection–di ffusion equation given by

∂ t i⫽ 冱 i

3 ∂ 3 c ∂ c ∂ 2 c ᎏ ᎏ⫹ V ᎏ ᎏ⫽ D ᎏ ᎏ⫹S

1 ∂ x i i⫽ 冱 1 ∂ x 2 i

PROPERTIES OF THE DIFFUSION EQUATION

where V i are the components of the ambient velocity and S m is the source flux of tracer per unit volume [M/L 3 T]. A reference concentration, C, such as the background concentration of the contaminant, can usually be de fined, along with a reference velocity, V, and a refer- ence length, L, which characterizes the dimension of the space in which the contaminant is moving. The concentration, c, coordinates, x i , time, t, and velocity components, V i , can

be normalized relative to these reference variables to yield the following nondimensional variables:

Substituting Equations 3.105 to 3.108 into Equation 3.104, taking S m ⫽ 0 (for a conserva- tive substance), and simplifying yields

i ∂ 2 c ᎏ * ᎏ⫹ 冱 V

i *ᎏ ᎏ⫽ 冱 ᎏ ᎏᎏ ᎏ

∂ t * i⫽ 1 ∂ x * i i⫽ 1 V L ∂ x * 2 i

The utility of this nondimensional representation is that all the terms involving nondimen- sional variables are on the order of one, since each of the nondimensional variables have been normalized by a reference quantity that is characteristic of the ambient environment. Consequently, the only terms whose magnitudes are not fixed are the diffusion terms, whose magnitudes are on the order of D i /VL. This nondimensional quantity represents the ratio of di ffusive transport to advective transport and is called the Peclet number, denoted by Pe. Therefore, de fining the Peclet number by

D Pe i ⫽ᎏ ᎏ i

Equation 3.109 can be written as

ᎏ ᎏ⫹ 冱 V *ᎏ i ᎏ⫽ 冱 Pe ᎏ ᎏ

According to Equation 3.111, whenever the component Peclet numbers, Pe i , are small, the di ffusion process can be neglected, and the contaminant transport is dominated by advec- tion. Conversely, for large Peclet numbers, di ffusion is the dominant process. The impor- tant result here is that the Peclet number is particularly useful as an indicator of the dominant transport process, which must be represented accurately in order to properly describe the transport of a contaminant in a particular environment.

In cases where the contaminant of interest is not conservative, the source term, S m , in the advection–di ffusion equation (Equation 3.104) is nonzero. In the special case where

FATE AND TRANSPORT IN AQUATIC SYSTEMS

the nonconservative contaminant exhibits first-order decay, the source term, S m , can be expressed as

(3.112) and the nondimensional advection–di ffusion equation can be written in the form

S m ⫽ ⫺kc

i⫽ 冱

V *ᎏ ᎏ⫽ Pe ᎏ

i i⫽ 冱 1 ∂ x * i 2 i

ᎏ ⫹ k*c*

The variable k* is the ratio of the relative magnitudes of the reaction rate and the advec- tion rate and is sometimes called the Damkohler number (Ramaswami et al., 2005) and denoted by “Da,” where

k L Da ⫽ ᎏ ᎏ

The Damkohler number is sometimes used to contrast the relative magnitudes of the reac- tion rate and di ffusion rate, in which case

Da i ⫽ᎏ ᎏ

where D i is the i-component of the di ffusion coefficient. For either definition of the Damkohler number, if Da ⬍⬍ 1, chemical reactions can be neglected.